Let \(\widehat{\mathbf {\theta }}_{obs\cdot em}\)  denote the maximizer of \(\mathbf {\beta }\) , \(\mathbf {\Sigma }_{\mathbf {\varepsilon }}\)  and \(\mathbf {\Sigma }_{\mathbf {b}}\)  in the observed data likelihood in (REF ). Similarly, let \(\widehat{\mathbf {\theta }}_{obs\cdot env}\)  denote the maximizer of \(\mathbf {\beta }\) , \(\mathbf {\Sigma }_{\mathbf {\varepsilon }}\)  and \(\mathbf {\Sigma }_{\mathbf {b}}\)  in (REF ) under additional conditions (i)\(^{*}\)  and (ii)\(^{*}\) . Let \(\mathbf {V}_0 = \mathrm {avar}(\widehat{\mathbf {\theta }}_{obs\cdot env})\)  and \(\mathbf {V} = \mathrm {avar}(\widehat{\mathbf {\theta }}_{obs\cdot em})\)  denote the asymptotic covariance matrices of the estimators obtained by directly maximizing (REF ) instead of using EM algorithm. Also, let \(\mathbf {\phi }= (\mathbf {\eta }, \mathbf {\Gamma }, \mathbf {\Omega }, \mathbf {\Omega }_0, \mathbf {\Sigma }_{\mathbf {b}})\)  and \({\mathbf {\theta }} = (\mathbf {\beta }, \mathbf {\Sigma }_{\mathbf {\varepsilon }}, \mathbf {\Sigma }_{\mathbf {b}})\)  denote the parameter under the envelope model and the standard model.
Let \(\lbrace \mathbf {\phi }_t\rbrace \)  and \(\lbrace {\mathbf {\theta }}_t\rbrace \)  denote the EM sequences, i.e., the parameters sequences we obtain from each EM iteration, of the envelope model and the standard model. By Corollary 1 of [1]}, the two EM sequences\( \lbrace \mathbf {\phi }_t\rbrace \)  and \(\lbrace {\mathbf {\theta }_t}\rbrace \)  converge to their unique maximizer of \(L_{obs}\) . Hence, in order to prove \(\mathbf {V}_{mix\cdot env}\le \mathbf {V}_{mix\cdot em}\) , it suffices to prove \(\mathbf {V}_0 \le \mathbf {V}\) . We found function \(\mathbf {h}\)  such that \(\mathbf {h}(\mathbf {\phi }) = (h_1(\mathbf {\phi }), h_2(\mathbf {\phi }), h_3(\mathbf {\phi }))\) . Because of the over-parameterization of \(\mathbf {\theta }\) , the gradient matrix \(\mathbf {G} = \dfrac{\partial \mathbf {h}(\mathbf {\phi })}{\partial \mathbf {\phi }^T}\)  is not of full rank. By Proposition 4.1 in [2]}, we have
\(\mathbf {V}_0 = \mathbf {G}(\mathbf {G}^T \mathbf {V}^{-1}\mathbf {G})^{\dagger }\mathbf {G}^T.\) 
\(\mathbf {V} - \mathbf {V}_0 = \mathbf {V} - \mathbf {G}(\mathbf {G}^T \mathbf {V}^{-1}\mathbf {G})^{\dagger }\mathbf {G}^T = \mathbf {V}^{\frac{1}{2}}\left[\mathbf {I} - \mathbf {V}^{-\frac{1}{2}} \mathbf {G}(\mathbf {G}^T \mathbf {V}^{-1}\mathbf {G})^{\dagger }\mathbf {G}^T \mathbf {V}^{-\frac{1}{2}}\right]\mathbf {V}^{\frac{1}{2}}.\) 
